Final Exam Study Guide and Practice Problems Solutions

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1 Final Exam Stuy Guie an Practice Problems Solutions Note: These problems are just some of the types of problems that might appear on the exam. However, to fully prepare for the exam, in aition to making sure you can o all of these problems, you shoul review your homework, quizzes, an class notes as well. Note problems ifferent than the ones liste here may appear on your exam. Section.: Introuction to Limits. The graph of f is shown below. Use the graph of f to evaluate the inicate its an function values. If a it or function value oes not exist, write DNE. a x 2 + = 4 b x 2 = 2 c DNE x 2 fx f 2 = e x fx = 2 f f2 = 2 g h fx = x 2 x 2 + fx = 2 i x 2 fx DNE 2. Suppose x fx = 3 an x gx = 2. Fin a x 3fx b gx fx x c x 3 fx 4gx 2 3. Fin each it a x 2 3 x 3 b 5 x = 5 c x x 2 = 2 x 2 + x x = 3 3 = 9 = 2 3 = = 6 6 = = = 9 3 = 6 = 2 + = 2 4. For each it, is the it a ineterminant form? Fin the it or explain why it oes not exist. x 7 2 a x 7 x 2 4x 2 x 2 b No; x 2 x + 2 Yes;

2 3x + 2 c x 4 x 2 4 x 2 9 x 3 x 3 No; DNE Yes; 6 5. Circle or to the following statements. fx a If fx = an gx =, then x x x gx oes not exist. fx b If fx = 2 an gx = 2, then x x x 2 gx = c In orer for fx to exist an be equal to 2, it must be the case that fx = 2 x x an fx = 2. x + If f is a function such that f2 exists, then x 2 fx exists. e If f is a polynomial, then x c fx = fc for every real number c. Section.2: Infinite Limits an Limits at Infinity. Suppose fx =. Fin each it. Write,, or DNE where appropriate. x 5 a fx x 5 Note the it will be ± since if we try plugging in x = 5 we get nonzero. To etermine which one, we plug in a number very close to 5 but to the left of 5, an we see zero that we get a negative number. Thus, fx =. x 5 b fx For similar reasons as in a, fx =. x 5 + x 5 + c fx DNE since the left an right-sie its are ifferent. x 5 2. Suppose fx = 3 x. Fin each it. Write,, or DNE where appropriate. x + 2 a b c fx x 2 Note the it will be ± since if we try plugging in x = 2 we get nonzero. To zero etermine which one, we plug in a number close to 2 but to the left of 2. For example, we can plug in 2. an we get that f 2. = 3 2. <. Thus, fx =. x 2 x 2 x 2 fx For similar reasons as in a, fx =. + + fx x 2 DNE since the left an right-sie its are not the same 2

3 3. Ientify all vertical an horizontal asymptotes of the following functions. a fx = x + 3 Note fx =. This is enough to show there is a vertical asymptote at x = 3. x 3 + There are no other zeros of the enominator other than x = 3, so this is the only vertical asymptote. Also, there is a horizontal asymptote at x = since x x + 3 =. b fx = x2 2x 3 x 2 9 x 3x + Note fx =. Note there is a hole in the graph at x = 3 since x = 3 is x + 3x 3 a zero of the enominator an yet x + fx = x 3 x 3 x + 3 = 4 6 So there is no vertical asymptote at x = 3. The only other zero of the enominator is x = 3. Note fx = x + x 3 + x 3 + x + 3 = an so there is a vertical asymptote at x = 3. Also, note that x 2 fx = x x x = 2 an so there is a horizontal asymptote at y =. 4. Fin the following its. Write or where appropriate. a x 2 2x x b x 3 x x x 2 2x = x 2 = x x c x 3 + 2x + x x x 3 x = x 3 = x x x 3 + 2x + x = x x x3 = 3

4 5. Circle or to the following statements. a A polynomial function of egree greater than or equal to has neither horizontal nor vertical asymptotes. b A rational function always has at least one vertical asymptote. c A rational function has at most one horizontal asymptote. 6. A theorem states that for n an a n a n x n + a n x n + + a = ± x What conitions must n an a n satisfy for the it to be +? Since a n x n + a n x n + + a = a n x n, only a n an x n etermine whether x x the it is or. If n is o, then x xn =. Thus, in this case, to make sure that a nx n =, we woul nee a n <. If n is even, then x x xn =, an so in this case we woul nee a n >. Thus, to summarize, we nee n o an a n <, or we nee n even an a n >. Section.5: Basic Differentiation Properties. If C is any constant, what is x C? C = for any constant C x 2. Write own the power rule. x xn = nx n for any real number n 3. If f x = 2x an g x = x 2, what is 2fx x 4 gx? 2fx x 4 gx = 2f x 4 g x = 22x 4 x2 = 4x 4 x2 4. Fin the erivative of the following functions. a fx = 7 f x = b fx = 2x + 3 f x = 2 + = 2 c fx = 2x 2 x + f x = 4x y = x + x 3/4 y x = 2 x / x /4 e y = 5x 2x 3 3 x 2/3 Note y = 5 x 3 2x x + 3 2/3 x = 2/3 5 x 3 2x 5/3 + 3x 2/3. Hence, y x = 3 5 x 4 3 x2/3 + 2x /3 Section.2: Derivatives of Exponential an Logarithmic Functions. Fin 5e x lnx + 2. x 5e x lnx + 2 = 5e x x x 4

5 2. Use log properties to fin erivatives of the following functions. Note: If you use the log properties you won t nee to use the chain rule. a fx = x 4 lnx 5 First, note fx = x 4 5 lnx. Thus, f x = 4x 3 5 x. b y = lnxe x First, note y = lnx + lne x = lnx + x. Thus, y x = x +. x c y = ln 5 x First, note y = lnx ln5 x = lnx ln5 + ln x = lnx ln5 2 lnx = lnx ln5 2 Alternatively, note y = ln 5 x/2 = ln Section.3: Derivatives of Proucts an Quotients. Fill in the blanks: lnx. Thus, y x = 2x. a The prouct rule says that given two ifferentiable functions F x an Sx, the erivative of their prouct is F xsx = F xsx + F xs x x b The quotient rule says that given two ifferentiable functions T x an Bx, the erivative of their quotient is T x T xbx T xb x = x Bx Bx 2 2. Fin x 2 e x. x x 2 e x = 2xe x + x 2 e x. x 3. If fx = x2 4x 3, fin f x. f x = 2x4x 3 x2 4 4x 3 2 = 4x2 6x 4x Fin f z if fz = z lnz. Do not simplify. f z = z z 3 z ln z z 3 z ln z z 3 z = z 3 2 ln z + z z 3 z ln z3z 2 z z Fin t e t t 2 3 lnt + t 3/2 e t t 2 = t 3 lnt + t 3/2. Do not simplify. = t e t t 2 3 ln t + t 3/2 e t t 2 t 3 ln t + t 3/2 3 ln t + t 3/2 2 e t t + 2 et t /2 3 ln t + t 3/2 e t t 2 3 t t/2 3 ln t + t 3/2 2 5

6 6. Fin f x if fx = ex lnx. Do not simplify. e x e x lnx e x f x = = x e x e x x e x 2 e x lnx e x 2 e x lnx e x x e x Section.4: The Chain Rule. Fin the inicate erivatives. You on t have to simplify. a f x if fx = 3x 2 2x f x = 3x 2 2x 9 6x 2 b y x if y = y lnx2 + x = x 2 + 2x c f w if fw = 3 2w ew 4 First, note that fw = 2w e w 4/3. Thus, f w = 4 3 2w ew 7/3 2 e w t 2 e t3 t t t 2 e t3 t = 2 t e t3 t + t 2 e t3 t 3t 2 t Section.5: Implicit Differentiation. Use implicit ifferentiation to fin y x if 2y xy2 + xy =. x 2y xy2 + xy = x 2y x y 2 + x2y y x y 2 2xy + x = y 2 y x y x = y 2 y 2 2xy + x + y + x y x = 2. Suppose e 2x y x 5 + x ln y =. Fin y x. e 2x y x 5 + x ln y = x x 2e2x 2x y y + e x 5x4 + ln y + x y y x = y e 2x + x = 5x 4 2e 2x y lny x y y x = 5x4 2e 2x y lny e 2x + x y 3. A curve is escribe by the equation y + x xy =. a Use implicit ifferentiation to fin y x. y + x xy = y + 2x + y + xy x x x x = y + x = 2x y x y x = 2x y + x 6

7 b Fin y. That is, fin y when x = an y =. x, x y = 2 = 3 x, + 2 c Fin the equation of the tangent line to the curve at,. The tangent line to the curve at, will go through, an have slope m = 3 2 which we foun in part b. Thus, the equation of the line can be foun by using the point-slope formula y y = mx x. Thus, the equation is y = 3 2 x y = 3 2 x y = 3 2 x Note you can leave the answer in either point-slope form or in slope-intercept form. Section 2.: First Derivative an Graphs. For each function, fin the intervals on which fx is increasing an the intervals on which fx is ecreasing. Fin any local maxes or mins. a fx = 2x 2 8x + 9 Note f x = 4x 8 = 4x 2. Thus, x = 2 is the only partition number for f x it is also a critical point for fx. We make a sign chart for f x. f x Thus, f is increasing on 2, an ecreasing on, 2. There is a local min at 2, f2 = 2,. There is no local max. b fx = x 4 + 5x 2 Note f x = 4x 3 + x = 4xx 2 25 = 4xx + 5x 5. Thus, the partition numbers of f x an the critical points of fx are x =, x = 5, an x = 5. The sign chart for f x is given below. f x Thus, we see that f is increasing on, 5 an on, 5 an ecreasing on 5, an 5,. There is a local max at 5, f 5 = 5, 625 an 5, f5 = 5, 625. There is a local min at,. c fx = x 3 + 3x 2 + 3x Hence the only parti- Note f x = 3x 2 + 6x + 3 = 3x 2 + 2x + = 3x + 2. tion number for f x is x =. Here is a sign chart for f x: f x Thus, f is increasing on,. There are no local maxes or mins. 7

8 2. Use the given info to sketch the graph of f. Assume its omain is,. ˆ f 2 = 4, f =, f2 = 4 ˆ f 2 =, f =, f 2 = ˆ f x > on, 2 an 2, ˆ f x < on 2, an, 2 Section 2:2: Secon Derivative an Graphs. Fin the inflection points of fx = lnx 2 4x + 5. This is a problem in your textbook. See the full solution on p The inflection points are, ln 2 an 3, ln Suppose fx = x /3. Fin the inflection points of fx. Also, fin the intervals on which fx is concave up an the intervals on which fx is concave own. Note f x = 3 x 2/3. So f x = 2 9 x 5/3 = Thus, the only partition number of f x is at x =. Here s a sign chart for f x5 x: f x Since f x changes sign at x = an since x = is in the omain of fx,, is an inflection point of fx. Also, notice f is concave up on, an concave own on,. 3. Suppose fx = x 4 2x 3. Fin a The omain of fx. b The intercepts of fx. c Make a sign chart for f x. Fin its critcal numbers. Fin any local extrema. List the intervals on which f is increasing an the intervals on which f is ecreasing. Make a sign chart for f x. Fin any inflection points an list the intervals on which fx is concave up or concave own. e Sketch the graph of f. The solution for this problem is given in your book on pages Section 2.3: L Hopital s Rule. For each it, etermine if the it is a ineterminate form, a ineterminate form, or neither. Can L Hopital s Rule be applie to fin the it? If not, explain why not. If it can be applie, use L Hopital s Rule to fin the it. ln + x 2 a Solution in book on p. 668 x x 4 ln x b Solution in book on p. 669 x x 8

9 c ln + 2e x x e x This is Matche Problem 8 in your book on p. 67. The solution is given on p. 674 e x x ln x Note that ln x = ; however, e x =. Thus, this is not a x x ineterminate form. Thus, L Hopital s Rule can not be use. e x 3x + e 3x x 2 Note that 3x + e 3x x = + e = = an x x 2 =. Thus, the it is a ineterminate form an hence L Hopital s Rule can be use. We get that 3x + e 3x x x 2 = x 3 3e 3x 2x This it is still a ineterminate form an we use L Hopital s Rule again to get that 3x + e 3x x x 2 9e 3x = x 2 = For each it liste, first a fin the it using previous methos, an then b use L Hopital s Rule to fin the it. a x 2x 2 x 2 + Using Theorem 4 on p. 59, we get that 2x 2 x x 2 + = 2x 2 x x = 2 2 Using L Hopital s Rule note the it is a ineterminate form, we get the same answer: b x 3x x 2 4 2x 2 x x 2 + = 4x x 2x = 2 Again we can use Theorem 4 on p. 59 to get 3x x x 2 4 = 3x x x = 3 2 x x = Using L Hopital s Rule note the it is a ineterminate form: c x 2 9 x 3 x + 3 3x x x 2 4 = 3 x 2x = We can evaluate this it using previous methos by factoring: x 2 9 x 3 x + 3 = x + 3x 3 x 3 x + 3 = x 3 = 3 3 = 6 x 3 Using L Hopital s Rule note the it is a ineterminate form: x 2 9 x 3 x + 3 = 2x x 3 = 2 3 = 6 9

10 Section 2.4: Curve Sketching Techniques. Use the graphing strategy taught in this section to analyze an graph the following functions. a gx = 4x + 3 x 2 This is Matche Problem 2 on p. 677 of your book. The solution is given on p b fx = xe.5x This is Matche Problem 3 in your book on p The solution is given on pages c fx = x ln x This is Matche Problem 4 in your book. The solution is given on p Sketch a function which satisfies the following properties. ˆ f 3 =, f =, f3 = ˆ f x < on, 2 an 2, ˆ f x > on 2, 2 ˆ f x < on, 2 an 2, ˆ f x > on, 2 an 2, ˆ vertical asymptotes: x = 2, x = 2 ˆ horizontal asymptotes: y = Section 2.5: Absolute Maxima an Minima. Fin the absolute maximum an absolute minimum of fx = x 3 2x on each of the following intervals: a [ 3, 3] Note f is continuous on [ 3, 3]. Thus, we just nee to evaluate f at any critical points an at the enpoints. The absolute max will be the maximum of these values, an the absolute min will be the minimum of these values. First, we fin the critical points. Note f x = 3x 2 2 = 3x 2 4 = 3x +2x 2. Thus, f x = at x = 2 an x = 2. The function values at these critical points are f 2 = = = 6 f2 = = 8 24 = 6

11 Now we evaluate f at the enpoints. Note f 3 = = = 9 f3 = = = 9 Thus, the absolute max is 6 an the absolute min is 6. b [ 3, ] Note f 3 = = = 9 f = 3 2 = We also know f evaluate at x = 2 from part a: f 2 = = = 6 Note the critical number x = 2 is outsie of the interval [ 4, ] so we on t care about f evaluate at this critical number. Thus, the absolute max is 6 an the absolute min is. 2. Fin the absolute extrema of each function on,. a fx = 2 x 5 x b fx = f lnx x This is Matche Problem 3 from your book. The answers are on p Section 2.6: Optimization. Fin the greatest possible prouct of two numbers given that the sum of the two numbers equals 28. Let x an y be the two numbers. We must have that x+y = 28. So y = 28 x. The function we want to maximize is fx = xy = x28 x = 28x x 2. Note f x = 28 2x. Thus, f x = when x = 4. A sign chart for f x woul show that there is an absolute maximum at x = 4. Thus, one of the numbers is 4. The other number is y = 28 x = 28 4 = 4. Thus, the two numbers are 4 an Fin two numbers whose ifference is 8 an whose prouct is a minimum. Let x an y be the two numbers. We must have x y = 8. So y = x 8. We want to maximize fx = xy = xx 8 = x 2 8x. Note f x = 2x 8. So f x = when x = 4. A sign chart for f x woul show that there is an absolute minimum at x = 4. Thus, one of the numbers is 4. The other number is y = x 8 = 4 8 = 4. Thus, the two numbers are 4 an Fin the area of the largest rectangle that can be mae with a perimeter of 56 ft. Let x an y be the lengths of the two sies of the rectangle. The perimeter is 56 ft, so we have the equation 2x + 2y = 56. This simplifies to x + y = 28. Thus, y = 28 x. We want to maximize Ax = xy = x28 x = 28x x 2. Note A x = 28 2x. Thus, A x = when x = 4. A sign chart woul show that there is an absolute max at x = 4. Thus, one of the sies has length 4 ft. Note when x = 4, y = 28 x = 28 4 = 4, so the other sie also has length 4 ft. This means the area woul be 4 2 = 96 ft Fin the largest possible perimeter of a rectangle whose area is 4 ft 2. Let x an y be the lengths of the two sies of the rectangle. The area of the rectangle is 4 ft 2, so we have the equation xy = 4. So y = 4 x. We want to maximize

12 4 P x = 2x + 2y = 2x + 2 = 2x + 8 x x. Note P x = 2 8 = 2x2 8. Thus, x 2 x 2 P x = when 2x 2 8 =. Note 2x 2 8 = 2x 2 4 = 2x 2x + 2 = x = 2 or x = 2 Since a length can t be negative, x = 2. A sign chart for P x woul show this is an absolute max. Thus, the largest possible perimeter occurs when x = 2. Note P 2 = = = 8. So the largest possible perimeter of a rectangle with area 4 2 ft 2 is 8 ft. 5. Suppose you want to fence a rectangular area with one sie against a barn so no fencing nees to be use for that sie. If the amount of fencing to be use is 4 ft, fin the imensions of the rectangle that has the maximum area. Drawing a picture here woul be a goo iea. Let x represent the length of each sie of the rectangle neither of which is opposite the barn. Let y represent the length of the sie of the rectangle which is opposite the barn. The amount of fencing to be use is 4 ft, so we must have that 2x + y = 4. So y = 4 2x. We want to maximize Ax = xy = x4 2x = 4x 2x 2. Note A x = 4 4x. So A x = when x =. A sign chart woul show that an absolute max occurs at x =. When x =, y = 4 2x = 4 2 = 2. Thus, the imensions are ft 2 ft with the sie of 2 ft of fencing being opposite the barn. Section 3.: Antierivatives an Inefinite Integrals. What oes it mean for F x to be an antierivative of fx? It means F x = fx. 2. If n is any real number, what o the family of antierivatives of x n look like? If n, then the family of antierivatives of x n look like n+ xn+ + C. If n =, then the family of antierivatives of x n = x look like ln x + C. 3. Fin an antierivative of fx = 3x 2. An antierivative of fx = 3x 2 is F x = x One coul also have chosen any function of the form F x = x 3 + C. 4. Fin an antierivative of fx = 2x. An antierivative of fx = 2x is F x = 2 ln x. One coul also have chosen any function of the form F x = 2 ln x + C. 5. Fin each inefinite integral. a x 2 5 x + 4ex x x 2 5 x + 4ex x = 3 x3 5 ln x + 4e x + C b c x 2 x x 2 x 2 x = x /2 2x 2 x = 2 x 2 3 x3/2 + 2x + C x x 5 x x x 5 x = x /2 x 5 x = 2x /2 6 x6 + C 2

13 xx 2 + x xx 2 + x = x 3 + x x = 4 x4 + 2 x2 + C One coul also o a u-substitution with u = x Circle or to the following statements. a Every function has an infinite number of antierivatives. b An antierivative of fx = x is F x = ln x +. c An antierivative of fx = 2x is x 2 + x. The constant function fx = is an antierivative of itself. e The function fx = 5e x is an antierivative of itself. f If n is an integer, then n + xn+ is an antierivative of x n. The integer coul be Section 3.2: Integration by Substitution. Use integration by substitution to fin each integral. a 2 x x Let u = 2 x. Then u = x an 2 x x = u /2 u = 2 3 u3/2 + C = x3/2 + C b x x 2 + x Let u = x 2 +, u = 2x x. Then x x 2 + x = u /2 u = u3/2 + C = x 2 + 3/2 + C 3 c x 2 x3 + 5 x Let u = x 3 + 5, u = 3x 2 x. Then x 2 x3 + 5 x = u 3 = u 3 u /2 u = 3 2u /2 + C = 2 3 x /2 + C 3

14 2. Integrate. a e 3x x x b x 2 9 x c 5t 2 t t This is Matche Problem 5 on p. 73. The answers are on p Section 3.4: The Definite Integral. Fin the area uner the graph of fx = x but above the x-axis between x = an x = 9. Note x lies completely above the x-axis between x = an x = 9. Thus, the area is given by 9 9 x x = x /2 x = x3/2 = /2 3/2 = = Consier the graph of fx shown below, where A, B, an C represent the shae regions shown. Suppose area of A =.5 area of B = 2.5 area of C =.5 Fin the following efinite integrals. a b c e fx x = B = 2.5 fx x = B C = = 3 fx x = A =.5 fx x = A B C = =.5 fx x fx x = A + B = = 5 4

15 3. Fin the following efinite integrals. a 2 x 3 x x 2 x 3 x x = 4 x4 2 2 x2 = = = = 9 4 b 3 6x 2 x x 3 6x 2 x = 2x 3 3 ln x x = 23 3 ln ln = 54 ln3 2 + = 52 ln3 c x x 2 2 x Let u = x 2 2, u = 2x x. Note when x =, u = 2 2 = 2 an when x =, u = 2 2 =. Thus, x x 2 2 x = 2 2 u u = 2 ln u = 2 2 ln ln 2 = 2 ln2 2 4 x x Let u = 4 x, u = x. When x =, u = 4 = 5. When x = 2, u = 4 2 = 2. Thus, x x = u u 5 = 2 3 u3/2 2 = /2 5 3/2 5

16 e 3 2x x2 x Let u = x 2, u = 2x x. x = 3, u = 3 2 = 8. Thus, Note when x =, u = 2 =, an when 3 2x 8 x2 x = u u = 8 u /2 u = 2u /2 8 = 2 8 = 2 8 = 22 2 = 4 2 Section 4.: Area Between Curves. Fin the area boune by the graphs of fx = x an gx = x 2 + 4x as shown below. The area is given by 3 x 2 + 4x x x = 3 x 2 + 3x x = 3 x x2 = = = = 9 2 6

17 2. Fin the area between the graph of fx = x 2 x 2 an the x-axis over [ 2, 2]. The graph of fx has been graphe below. The area will be the sum of the areas of the two shae regions A an B as shown. We nee to evaluate two separate integrals here, one to fin the area of region A an one to fin the area of region B. We can then a these together to get the total area. First, we fin the area of region A. This is given by 2 x 2 x 2 x = 3 x3 2 x2 2x 2 = = = = = = 6 Now we fin the area of region B. This is given by 2 x 2 x 2 2 x = x 2 + x + 2 x = 3 x x2 + 2x = = = = = 5 2 = 9 2 Thus, the total area is =

18 Section 5.: Functions of Several Variables. Given the functions fx, y = x 2 xy + 2y an gx, y = x2 2 y +, fin a f, f, = = 2 b f, 2 f, 2 = =; c g2, 3 g2, 3 = = 2 4 = 2 g, 2 g, 2 = = 3 e g2, Note the enominator is zero when y =. Thus, g2, is not efine. f f2, 2 g2, 2 f2, 2 g2, 2 = = = 3 Section 5.2: Partial Derivatives. Given the functions fx, y = x 2 xy + 2y an gx, y = x2 2, fin the inicate partial y + erivatives. a f x b f y c f xx f yx e g x f g y g g xy h g yy a f x = 2x y b f y = x + 2 c f xx = 2 f yx = e g x = 2x y + f Note gx, y = x 2 2y +. So g g xy = 2x y + 2 h g yy = x y + 3 = 2x2 4 y + 3 g y = x 2 2 y + 2 = x2 + 2 y + 2 8